# Cos x and Sin x

1. Mar 11, 2007

### jackson_sun

I have been asked to differentiate cos x and six....the maclaurin series versions...

I have done the general and specific terms as shown below.

Im not sure if these are correct?
thanks

General Terms
cos x = ∑ (-1)n (x^(2n) / (2n)!)

COS x = ∑ (-1)n (x^(2n+1)/ (2n +1) / (2n)!)

sin x = ∑ (-1)n (x^(2n+1) / (2n+1)!)

‘sin x = ∑ (-1)n (((2n+1) x^(2n))/ (2n+1)!)

Specific Terms
sin x = x – (x^3 / 3!) + (x^5 / 5!) – (x^7 / 7!) + (x^9 / 9!) - …
‘sin x = 1 – (3x^2 / 3!) + (5x^4 / 5!) - (7x^6 / 7!) + (9x^8 / 9!) - …

cos x = 1 – (x^2 / 2!) + (x^4 / 4!) – (x^6 / 6!) + (x^8 / 8!) - …
COS x = x – (x^3 / 3 / 2!) + (x^5 / 5 / 4!) – (x^7 / 7 / 6!) + (x^9 / 9 / 8!) - …
thanks

2. Mar 11, 2007

### Gib Z

You already posted the same thing else where, learn what the factorial actually means! Then you can simplify yours derivatives and realize what your not realizing right now!

3. Mar 11, 2007

### HallsofIvy

You know to use "^" to indicate exponent of the "x"- use for the exponent on (-1) also: (-1)^n. Obviously "cos x" can't be equal to both of these. Do you mean cos' x? Also A/B/C, at best, ambiguous. Do you mean
((x^(2n+1)/(2n+1))/(2n)! or (x^(2n+1))/((2n+1)/(2n)!)?

Finally, if you are differentiating as you say, you are "going the wrong way"! The derivative of x^n is nx^(n-1), not (1/(n+1)) x^(n+1).

4. Mar 11, 2007

### Gib Z

It looks to me the cos x in capitals is the integral of cos x in normal...

5. Mar 12, 2007

### jackson_sun

yeh COS x is the integral of cos x....i meant integral of cos x and derivative of sin x...sorry for the confusion.

I have realised how to simplify 'sin x so that it equals cos x....but i am not sure how to simplify the integral of cos x to equal sin x.

COS x = x – (x^3 / 3 / 2!) + (x^5 / 5 / 4!) – (x^7 / 7 / 6!) + (x^9 / 9 / 8!)
= ?

any idea? and on the subject...what is the notation for the integral of cos x...I have used COS x but this is obviously incorrect.
thanks...

6. Mar 12, 2007

### Gib Z

Take your second term...if you don't see the arithmetic, don't try this level of math yet, $$\frac{\frac{x^3}{3}}{2!} =\frac{x^3}{3\cdot2!} = \frac{x^3}{3!}$$.

I didn't take into account the sign of the integral, but you get what i mean.

To learn how to type the math graphics im typing, click on the images I make appear. That will tell you what i typed to get that. Or at least say, "Integral of.."...

7. Mar 12, 2007

### jackson_sun

I understand exactly what you have done.
How would it be done for the general terms is my bigger problem

8. Mar 12, 2007

### Gib Z

Good
$$\cos x = \sum_{n=0}^{\infty} -1^n \frac{x^{2n}}{2n!}$$

Integrate every term in the sum, with respect to x. In every individual term, n is constant.

So

$$\int\cos x dx= \sum_{n=0}^{\infty} -1^n \frac{\frac{x^{2n+1}}{2n+1}}{2n!}$$

(2n+1)2n!=(2n+1)!,

therefore it simplifies to

$$\int\cos x dx= \sum_{n=0}^{\infty} -1^n \frac{x^{2n+1}}{(2n+1)!}$$, which is exactly your series for sin x.

9. Mar 12, 2007

### HallsofIvy

Just one comment: it would be far better to write
$$\int\cos x dx= \sum_{n=0}^{\infty} (-1)^n \frac{x^{2n+1}}{(2n+1)!}$$
with parentheses around the -1. I would be inclined to interpret -1n as -(1n) which, of course, would be just -1!

10. Mar 12, 2007

### Gib Z

O yes i guess i would too, if someone else wrote that..thanks

11. Mar 12, 2007

### jackson_sun

Thanks gents

12. Mar 12, 2007

### Gib Z

No Problem, and welcome to PF